Answer:
option D
Step-by-step explanation:
Clearly from the graph we say that at x = 4 and x = -6, f(x) = 0
Now we will check this with your options.
A) f(x) = (x + 1 )² - 5
at x = 4, f(x) = 25 - 5 = 20 ≠ 0 ; not A
B) f(x) = x - 5
at x = 4, f(x) = 4 - 5 = 1 ≠ 0 ; not B
C) f(x) = (x + 1 )³ - 5
at x = 4 , f(x) = 125 - 5 = 120 ≠ 0 ; not C
D) f(x) = | x + 1 | - 5
at x = 4 , f(x) = | 4 + 1 | - 5
= 5 - 5 = 0 ; satisfies
at x = -6 , f(x) = | - 6 + 1 | - 5
= | - 5 | - 5
= 5 - 5 = 0 ; D satisfies [ | - a | =a , absolute value function ]
